The collision between two pucks

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SUMMARY

The discussion focuses on solving a two-dimensional collision problem involving two pucks, A and B, with masses of 0.294 kg and 0.588 kg, respectively. Puck A moves at an initial velocity of 5.55 m/s along the x-axis, while puck B is initially at rest. The final speeds calculated for puck A and puck B after the collision are 3.41 m/s and 2.57 m/s, respectively. The solution involves applying conservation of momentum in both the x and y components, using the equations m1v0 = m1v1' cos θ1 + m2v2' cos θ2 and 0 = m1v1' sin θ1 - m2v2' sin θ2.

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Homework Statement


Puck A has a mass of 0.294 kg and is moving along the x-axis with a velocity of 5.55 m/s. It makes a collision with puck B, which has a mass of 0.588 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of
Puck A
Puck B

Homework Equations


I have spent an hour on this problem. I have two different eqns one for x and one for y, respectively:
m1v0 = m1v1 cos 65 + m2v2cos 37
and
0 = m1v1 sin 65 - m2v2 sin 37



The Attempt at a Solution


I solved for v2 from the eqn for y and tried to plug that value into the eqn for x. This did not work and I am running out of ideas and patience. Please help! There is also a pic for this problem..
 

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I think you accidentally messed up you arithmetic when solving the question. Your X and Y equations are correct so you've set up your question correctly. i think you have just made a mistake when solving because the substitution of V2' of you Y equation into your X equation is the correct thing to do.

I got V1' = 3.41 m/s; V2' = 2.57 m/s; And these numbers make sense in my mind
 
Your numbers worked out perfectly. I don't understand where i am going wrong, because i am not getting your numbers. can you do a step by step please? id really appreciate it.
 
X-component: m1v1 = m1v1' cos 65 + m2v2'cos 37
Y-Component: 0 = m1v1' sin 65 - m2v2' sin 37

Y: 0 = m1v1' sin 65 - m2v2' sin 37
m1v1' sin 65 = m2v2' sin 37
m1v1' sin65/(m2sin37) = v2' -> Plug in numbers
(0.294 kg)*(v1')*(sin65)/[(0.588 kg)(sin37)] = v2'
0.75298*v1' = v2' -> substitute into the X

m1v1 = m1v1' cos 65 + m2v2'cos 37
(0.294 kg)(5.55 m/s) = (0.294 kg)v1'(cos65) + (0.588 kg)(0.75298v1')(cos37)
1.6317 kg*m/s = 0.12425*v1' kg + 0.3536*v1' kg
1.6317 kg*m/s = 0.4778*v1' kg
v1' = 3.41 m/s -> substiture back into Y

0.75298*v1' = v2'
0.75298*(3.41 m/s) = v2'
v2' = 2.57 m/s

So this was my solution... a lot messier on the computer but i hope you can find where you and i differed in our solutions and this can help you avoid the same mistake in the future.

Cheers
 
Thankyou so much! This was such a help to say the least
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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